## How does the concentration change the equilibrium constant?

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### How does the concentration change the equilibrium constant?

I thought I remembered Dr Lavelle saying that the amount of product / reactant doesn't affect a reaction's equilibrium constant if the conditions are the same. But in question 1 in topic 5H we have to calculate the K for variations of the same reaction, and each has a different K. I understand the inverse reaction's logic; but I don't see how changing the molarity of all substances results in a different K chemically speaking (i.e. I know you change exponents but I don't see why we have different K if they're the same reaction)

Question 5H.1: For the reaction N2(g) + 3 H2(g) <=> 2 NH3(g) at 400. K, K = 41. Find the value of K for each of the following reactions at the same temperature:
b) 1 2 N2(g) + 3 2H2(g) <=> NH3(g)
c) 2 N2(g) + 6 H2(g) <=> 4 NH3(g)

Ryan Narisma 4G
Posts: 104
Joined: Fri Aug 30, 2019 12:18 am

### Re: How does the concentration change the equilibrium constant?

Hi nicolely3B! You are correct in saying that the K doesn't change as long as it is the exact same reaction. In the practice problems, these reactions are not exactly the same. They differ by the stoichiometric coefficients used in the Keq expression. The original expression looks something like this:
Keq= = 41

In part b, the equilibrium expression looks like so:
Keq=

Notice how the equilibrium expression in part b looks very similar to the original expression. The only difference is that it seems like everything in the expression was raised to the power of (1/2). Thus, the value of K of the reaction in b appears to be (41)^(1/2).

So, K would be the same if the reaction is exactly the same and at the same temperature.

I hope this explains it a bit!

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### Re: How does the concentration change the equilibrium constant?

Ryan Narisma 4G wrote:Hi nicolely3B! You are correct in saying that the K doesn't change as long as it is the exact same reaction. In the practice problems, these reactions are not exactly the same. They differ by the stoichiometric coefficients used in the Keq expression. The original expression looks something like this:
Keq= = 41

In part b, the equilibrium expression looks like so:
Keq=

Notice how the equilibrium expression in part b looks very similar to the original expression. The only difference is that it seems like everything in the expression was raised to the power of (1/2). Thus, the value of K of the reaction in b appears to be (41)^(1/2).

So, K would be the same if the reaction is exactly the same and at the same temperature.

I hope this explains it a bit!

It explained everything, thanks a lot!